1014. Best Sightseeing Pair**
https://leetcode.com/problems/best-sightseeing-pair/
题目描述
Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.
The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.
Return the maximum score of a pair of sightseeing spots.
Example 1:
Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11
Note:
-
2 <= A.length <= 50000 -
1 <= A[i] <= 1000
C++ 实现 1
参考 Detailed Explanation using DP [O(n) Time | O(1) Space], 求和分为两个部分 (A[i] + i) 以及 A[j] - j. 使用 res 来记录 (A[i] + A[j] + i - j) 的最大值, 使用 max_prev 来不断更新 A[i] + i 的值. 由于 max_prev 保存 A[i] + i 的值, 因此更新 res 时, 使用 max_prev + A[j] - j.
class Solution {
public:
int maxScoreSightseeingPair(vector<int>& A) {
int res = 0, max_prev = 0;
for (int i = 0; i < A.size(); ++ i) {
res = std::max(res, max_prev + A[i] - i);
max_prev = std::max(max_prev, A[i] + i);
}
return res;
}
};
C++ 实现 2
思路和 C++ 实现 1 一致.
class Solution {
public:
int maxScoreSightseeingPair(vector<int>& A) {
int res = 0, max_prev = A[0] + 0;
for (int i = 1; i < A.size(); ++ i) {
max_prev = std::max(max_prev, A[i - 1] + i - 1);
res = std::max(res, max_prev + A[i] - i);
}
return res;
}
};
